What is the answer to solvefor r,x^2(r^2-x^2)^{0.5}=1 ?

The answer to solvefor r,x^2(r^2-x^2)^{0.5}=1 is r=(sqrt(x^6+1))/(x^2),r=-(sqrt(x^6+1))/(x^2)

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solvefor r,x^2(r^2-x^2)^{0.5}=1

solve for

r, x^{2} (r^{2} - x^{2})^{0.5} = 1

r = \frac{x ^{6} + 1}{x ^{2}}, r = - \frac{x ^{6} + 1}{x ^{2}}

Solution steps

x^{2} (r^{2} - x^{2})^{0.5} = 1

Divide both sides by

x^{2}

(r^{2} - x^{2})^{0.5} = \frac{1}{x ^{2}}

Square both sides:

r^{2} - x^{2} = \frac{1}{x ^{4}}

r^{2} - x^{2} = \frac{1}{x ^{4}}

Solve

r^{2} - x^{2} = \frac{1}{x ^{4}} : r = \frac{x ^{6} + 1}{x ^{2}}, r = - \frac{x ^{6} + 1}{x ^{2}}

r = \frac{x ^{6} + 1}{x ^{2}}, r = - \frac{x ^{6} + 1}{x ^{2}}

Verify Solutions:

r = \frac{x ^{6} + 1}{x ^{2}}

True

, r = - \frac{x ^{6} + 1}{x ^{2}}

True

The solutions are

r = \frac{x ^{6} + 1}{x ^{2}}, r = - \frac{x ^{6} + 1}{x ^{2}}

x^{2} + 4 x = x^{2} + 4 x

2 x - 3 + 2 x + 4 = 6 x + 13

x^{0.4} = 8

3 x - 5 = x - 1

5 x^{\frac{1}{2}} = 2 x + 2

What is the answer to solvefor r,x^2(r^2-x^2)^{0.5}=1 ?
The answer to solvefor r,x^2(r^2-x^2)^{0.5}=1 is r=(sqrt(x^6+1))/(x^2),r=-(sqrt(x^6+1))/(x^2)